# $A \subset B$ be integral and flat extension then it is faithfully flat.

I want to prove the following:

$$A \subset B$$ be integral and flat extension of rings then it is faithfully flat.

Clearly enough to show that for every ideal $$I$$ of $$A$$, $$I^{ec}=I$$. Since the extension is integral for every prime ideal $$p$$ of $$A$$ $$p^{ec}=p$$. I can’t show that it also holds for arbitrary ideal. I need some help. Thanks.

It is enough to prove that for any prime ideal $$\mathfrak p\in\operatorname{Spec} A$$, there exists a prime ideal $$\mathfrak q\in\operatorname{Spec} B$$ such that $$\;\mathfrak q\cap A=\mathfrak p$$.